Divisors - Elliptic Curves: Number Theory and Cryptography - page 376

LEMMA 11.27
A ssu m e that the pairing

, : B × A → μ n is nondegeneratein A (that is, if

b, a =1 for all b ∈ B ,then a =0 ).

(a) T he m ap A

ψ a isinjective.
(b) If # A =# B ,then the pairing isalso nondegeneratein B .

→ Hom( B, μ n ) given by a →

PROOF Suppose ψ a is the trivial homomorphism. This means that b, a =

ψ a ( b ) = 1 for all b ∈ B . The nondegeneracy in A implies that a =0. This

proves (a).
Let

B 1 = {b ∈ B |b, a =1forall a ∈ A}.

Then each a ∈ A gives a well-defined homomorphism β a : B/B 1 → μ n
given by β a ( b mod B 1 )= b, a . f β a is the trivial homomorphism, then
b, a =1forall b ∈ B , which means that a = 0. Therefore, A injects
into Hom( B/B 1 ,μ n ), which has order # B/ # B 1 , by Lemma 11.26. Since
# A =# B ,wemusthave# B 1 =1. But B 1 = 0 is exactly what it means for
the pairing to be nondegenerate in B . This proves (b).

A converse of part (b) of Lemma 11.27 holds.

LEMMA 11.28
Suppose

, : B × A → μ n is nondegenerateinboth A and B .Then # A =

# B .Infact, A Hom( B, μ n ) and B Hom( A, μ n ) .

PROOF By Lemma 11.27, we have an injection from A to Hom( B, μ n ),

so # A ≤ #Hom( B, μ n )=# B . Reversing the roles of A and B ,wehave

# B #Hom( A, μ n )=# A . Therefore, # A =# B , and the injections are

isomorphisms.

≤

LEMMA 11.29
Let M be a finiteabe ian group and let α : M

→ M be a hom om orphism .

Then

#Ker α =# M/ # α ( M ) .

PROOF By Theorem B.6 (in Appendix B),

# M =(#Ker α )(# α ( M )) .

The result follows.

The following technical lemma is the key to proving the nondegeneracy of
the Tate-Lichtenbaum pairing.